Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 7x}{x - 1} = \dfrac{-17x - 9}{x - 1}$
Solution: Multiply both sides by $x - 1$ $ \dfrac{x^2 - 7x}{x - 1} (x - 1) = \dfrac{-17x - 9}{x - 1} (x - 1)$ $ x^2 - 7x = -17x - 9$ Subtract $-17x - 9$ from both sides: $ x^2 - 7x - (-17x - 9) = -17x - 9 - (-17x - 9)$ $ x^2 - 7x + 17x + 9 = 0$ $ x^2 + 10x + 9 = 0$ Factor the expression: $ (x + 1)(x + 9) = 0$ Therefore $x = -1$ or $x = -9$ The original expression is defined at $x = -1$ and $x = -9$, so there are no extraneous solutions.